Solution Key
Name:
PHYS 3410/6750: Modern Optics
Final Exam
Friday 14 December 2012
Prof. Bolton
No books, calculators, notes, etc.
Formulas of possible interest
D
I = 0 c E 2
E
T
1
= 0 cE02
2
Eγ = hνγ
n = c/v
v = λν
(v = “vee”; ν = “nu”.)
k = 2π/λ
ω = 2πν
τ = 1/ν
D=
1
1
1
= (nl − 1)
−
f
R1 R2
1
1
1
+ =
so si
f
si
yi
=−
MT =
yo
so
n1 sin θ1 = n2 sin θ2
θr = θi
a(sin θm − sin θi ) = mλ
P = σAT 4
λpeak T = const.
eiφ = cos φ + i sin φ
PHYS 3410/6750
Final Exam with solutions, Fall 2012 2
Problem 1 (25 points)
If you look at a flat mirror, you will see a virtual image of yourself on the other side that
should be familiar. If you look at a convex mirror, you will see a demagnified virtual image
of yourself on the other side. This is depicted qualitatively in the following figure:
you
you
flat mirror
your
mirror
image
convex mirror
your
mirror
image
The paraxial lens equation for a spherical mirror is
1
1
2
+ =−
so si
R
where so is the object distance, si is the image distance, and R is the radius of curvature
of the mirror. By convention, si is negative for virtual images and positive for real images,
while R is negative for concave mirrors and positive for convex mirrors.
Using this equation, derive an expression for the relative angular size between your image
as seen by you when standing a distance d from a convex mirror with radius of curvature
R, and your image as seen by you when standing a distance d from a flat mirror. Your
expression should depend only on the ratio d/R. (Although you will most likely want to
define additional variables for use during your derivation.)
To do this, first find a general expression for the total apparent distance from you to your
mirror image, then find an expression for the apparent linear height of your mirror image.
The ratio of the latter to the former gives the apparent angular size of your image (in the
small-angle approximation, which you may use). Finally, take the ratio of the apparent
angular sizes for the convex-mirror case to the flat-mirror case to obtain your result.
Consider the convex mirror case first. Solving the mirror equation and putting in d (your
distance from the mirror) for the object distance, we get:
1
2
1
=− −
si
R d
Since R is by convention positive for a convex mirror and d is also positive, 1/si will be
negative, and so will si . This just confirms that the image is virtual. Let us define x ≡ −si
to be the apparent distance of the virtual image “behind” the mirror. This gives
1
2
1
= +
x
R d
⇒ x=
Rd
.
R + 2d
PHYS 3410/6750
Final Exam with solutions, Fall 2012 3
The apparent distance from you to your image is then
d+x=d+
Rd
.
R + 2d
The apparent linear image height is
yi = −
si
x
Ryo
yo = yo =
so
d
R + 2d
where yo is the object height—that is, your actual height.
The angular size of your image (call it αR ) is then
Ry
o
Ryo
yo
yi
R+2d
αR =
=
=
Rd =
d+x
d(R + 2d) + Rd
2d(1 + d/R)
d + R+2d
A flat mirror corresponds to R = ∞, so the angular size of your image in the flat mirror will
be
yo
yo
=
α∞ =
2d(1 + d/∞)
2d
Finally, the relative angular image size for a convex mirror to a flat mirror is
αR
yo
=
α∞
2d(1 + d/R)
2d
yo
!
=
1
.
1 + d/R
PHYS 3410/6750
Final Exam with solutions, Fall 2012 4
Problem 2 (25 points)
The far-field (Fraunhofer) diffraction pattern of an aperture in terms of angular coordinates
θx and θy can be found from
I(θx , θy ) ∝ |T (kθx , kθy )|2
where the function T (kx , ky ) is the two-dimensional Fourier transform of the amplitude
transmission coefficient t(x, y) over the aperture:
T (kx , ky ) =
Z +∞
dx eikx x
Z +∞
dy eiky y t(x, y)
−∞
−∞
Use this to find the Fraunhofer diffraction pattern I(θx , θy ) of a square aperture of side
length a. (That is, t(x, y) = 1 when −a/2 < x < +a/2 and −a/2 < y < +a/2, and t(x, y) = 0
otherwise.) You may neglect any overall constants of proportionality in your result—only
the functional dependence of I on θx and θy is of interest. Please express your final result
entirely in terms of real numbers, also using the wavelength of light λ instead of the angular
spatial frequency (or wavenumber) k.
For our square aperture, the Fourier transform integral becomes
T (kx , ky ) =
Z +a/2
dx eikx x
−a/2
Z +a/2
dy eiky y
−a/2
The x and y integrals are separable. The x integral evaluates to
+a/2
eikx x eikx a/2 − e−ikx a/2
sin(kx a/2)
=
∝
ikx x=−a/2
ikx
kx
and the y integral evaluates similarly, so that
T (kx , ky ) ∝
sin(kx a/2) sin(ky a/2)
kx ky
Substituting in with kx = kθx , ky = kθy , and k = 2π/λ, and taking the squared modulus to
get the irradiance pattern, we obtain
I(θx , θy ) ∝
sin2 (πθx a/λ) sin2 (πθy a/λ)
,
θx2 θy2
where we have ignored any pre-factors that do not affect the θx , θy dependence.
PHYS 3410/6750
Final Exam with solutions, Fall 2012 5
Problem 3 (25 points)
The electric field amplitude within a waisted Gaussian laser beam has the form
"
E(r, z) = E0
−r2
w0
exp
w(z)
[w(z)]2
#
(
)
where w0 is the width of the beam at its waist, z is the distance from the waist along the
beam in the direction of propagation, r is the perpendicular distance outward from the axis
of the beam, and w(z) is the z-dependent beam width parameter given by
s
w(z) = w0
z
1+
zR
2
.
The Rayleigh range zR is given by
πw02
.
λ
For a laser of this type with wavelength λ = 10−6 m and beam waist w0 = 10−3 m, find the
value of z at which the irradiance I at the center of the beam (i.e., at r = 0) drops to 1%
of its value at z = 0. One significant figure is sufficient in your answer. Assume that the
z value of interest is sufficiently greater than zR to employ an approximation to w(z), then
verify that the value you find justifies this assumption to the requested level of accuracy.
zR =
Irradiance is proportional to electric field squared. Since we are ultimately concerned with a
ratio of irradiances, write I = E 2 for convenience, ignoring the constants of proportionality.
At z = 0, w(z) = w0 , and so we simply have I(0, 0) = E02 . For z significantly larger than zR ,
we have
s z 2 w0 z
w0 zλ
zλ
=
=
=
.
w(z) ' w0
2
zR
zR
πw0
πw0
This gives
"
I(0, z) =
E02
w0
w(z)
#2
=
E02
π 2 w04
z 2 λ2
!
We want to find z such that I(0, z)/I(0, 0) = 1/100, which corresponds to
1
π 2 w04
=
2
2
z λ
100
⇒ z = 10 ×
πw02
(10−3 m)2
= 10 × 3.14 ×
' 30 m.
λ
10−6 m
Since this is 10× the Rayleigh range zR , the approximation only incurs 1% errors in [w(z)]2
and I(0, z), which is sufficient for our level of accuracy.
PHYS 3410/6750
Final Exam with solutions, Fall 2012 6
Problem 4 (25 points)
The wire filament of a 100-watt incandescent light bulb operates by radiating thermally at
a temperature of 2,000 K.
(a) To the nearest order of magnitude, what is the effective surface area of the filament in
square millimeters? Take the Stefan-Boltzmann constant to be σ = 6 × 10−8 W m−2 K−4 .
(b) If the circumference of the filament wire is 0.2 millimeters, approximately what length of
wire is necessary to make the filament. Assume that your order-of-magnitude answer from
part (a) is accurate to one significant figure. How could this length be fit inside a common
light bulb?
(c) At what wavelength does the emission spectrum of the filament peak? Take the constant
in Wien’s displacement law to have the value 3 × 10−3 m·K exactly.
(d) Why is this light bulb an energy-inefficient source of visible light?
(a) Using the Stefan-Boltzmann law P = σAT 4 we have
A=
(100 W)
P
≈
≈ 10−4 m2 = 100 mm2 .
4
−8
−2
−4
3
4
σT
(6 × 10 W m K )(2 × 10 K)
(Note that 6 × 24 = 6 × 16 ≈ 100.)
(b) If L is the wire length and C is the circumference, then the suface area is A = CL. So,
L = A/C =
1
100 mm2
= 500 mm = m
0.2 mm
2
This is enclosed within a lightbulb by making it into a coil. (Actually, a coiled coil!)
(c) Using Wien’s displacement law λpeak T = const. we have
λpeak =
const.
3 × 10−3 m · K
=
= 1.5 × 10−6 m.
T
2 × 103 K
(d) This is an energy-inefficient source of visible light because its radiation spectrum peaks
outside of the visible wavelength range (at infrared wavelengths).